Adding randomization to simple testimonial rotator

Question

There was an awesome, super-minimal testimonial rotator that I found here, but I'm curious as to how I might get it to simply randomize the results. Here's the rotator as it now functions:

$('.testimonials div:first').show();

setInterval(function(){ 
    $('.testimonials div:first-child')
        .fadeOut(1000)
        .next('div')
        .delay(1000)
        .fadeIn(1000)
        .end()
        .appendTo('.testimonials') 
},3000);

http://jsfiddle.net/xXRwA/9/

Answer 1

Try this:

var $items = $('.testimonials .item');

function getRandomItem(){
    return $items.eq(Math.floor($items.length * Math.random()));
}

getRandomItem().show();

setInterval(function(){ 
    var $outgoing = $items.filter(':visible');
    var $incoming = getRandomItem();
    $outgoing.fadeOut(1000, function(){
        $incoming.fadeIn(1000);
    });
}, 3000);

Demo: http://jsfiddle.net/JWGbz/6/

Answer 2

This code makes sure the same item isn't shown twice and continues to work if you change the amount of testimonials. The first item shown is also random.

Demo

$(document).ready(function() {

    $('.testimonials div').eq(Math.floor((Math.random() * $('.testimonials div').length))).show();

    setInterval(function() {

        var $items = $('.testimonials div');
        if ($items.length < 2) {
            return;
        }

        var $currentItem = $items.filter(':visible');
        var currentIndex = $currentItem.index();
        var nextIndex = currentIndex;
        while (nextIndex == currentIndex) {
            nextIndex = Math.floor((Math.random() * $items.length));
        }
        $currentItem.fadeOut(1000, function() {
            $items.eq(nextIndex).fadeIn(1000);
        });

    }, 3000);

});

Answer 3

This problem kept eating at me, and I realized that the real problem is here is that you want a way to shuffle the testimonials. If you had a way to do that, then your original function would even work as intended. As it turns out, shuffling a list of jQuery elements is not as easy as you might think. I started by implementing a function that allows you to swap two arbitrary jQuery elements (this avoids using jQuery.clone, which has side-effects, like removing event listeners):

function swap($a, $b){
    var $aNext = $a.next(),
        $aParent = $a.parent();
    $a.insertAfter($b);
    if($aNext.length) $b.insertBefore($aNext);
    else $aParent.append($b);
}

Then you can implement a Fisher-Yates shuffle:

function shuffle($items){
    var i, j;
    for(i=$items.length-1; i>1; i--){
        j = Math.floor(Math.random() * (i+1));
        swap($items.eq(i), $items.eq(j));
    }
}

Now you can simply shuffle all of your testimonials:

shuffle($('.testimonials .item'));

Then use your original code:

$('.testimonials div:first').show();

setInterval(function(){ 
    $('.testimonials div:first-child')
        .fadeOut(1000)
        .next('div')
        .delay(1000)
        .fadeIn(1000)
        .end()
        .appendTo('.testimonials') 
},3000);

Of course, once you've gone through all the testimonials, you may want to re-shuffle them to it doesn't keep repeating the same order over and over again.

Demo: http://jsfiddle.net/JWGbz/7/